Quotient by subgroup is well defined if and only if subgroup is normal

Let \(G\) be a group and \(N\) a nor­mal sub­group of \(G\). Then we may define the quo­tient group \(G/N\) to be the set of left cosets \(gN\) of \(N\) in \(G\), with the group op­er­a­tion that \(gN + hN = (gh)N\). This is well-defined if and only if \(N\) is nor­mal.


\(N\) nor­mal im­plies \(G/N\) well-defined

Re­call that \(G/N\) is well-defined if “it doesn’t mat­ter which way we rep­re­sent a coset”: whichever coset rep­re­sen­ta­tives we use, we get the same an­swer.

Sup­pose \(N\) is a nor­mal sub­group of \(G\). We need to show that given two rep­re­sen­ta­tives \(g_1 N = g_2 N\) of a coset, and given rep­re­sen­ta­tives \(h_1 N = h_2 N\) of an­other coset, that \((g_1 h_1) N = (g_2 h_2)N\).

So given an el­e­ment of \(g_1 h_1 N\), we need to show it is in \(g_2 h_2 N\), and vice versa.

Let \(g_1 h_1 n \in g_1 h_1 N\); we need to show that \(h_2^{-1} g_2^{-1} g_1 h_1 n \in N\), or equiv­a­lently that \(h_2^{-1} g_2^{-1} g_1 h_1 \in N\).

But \(g_2^{-1} g_1 \in N\) be­cause \(g_1 N = g_2 N\); let \(g_2^{-1} g_1 = m\). Similarly \(h_2^{-1} h_1 \in N\) be­cause \(h_1 N = h_2 N\); let \(h_2^{-1} h_1 = p\).

Then we need to show that \(h_2^{-1} m h_1 \in N\), or equiv­a­lently that \(p h_1^{-1} m h_1 \in N\).

Since \(N\) is closed un­der con­ju­ga­tion and \(m \in N\), we must have that \(h_1^{-1} m h_1 \in N\); and since \(p \in N\) and \(N\) is closed un­der mul­ti­pli­ca­tion, we must have \(p h_1^{-1} m h_1 \in N\) as re­quired.

\(G/N\) well-defined im­plies \(N\) normal

Fix \(h \in G\), and con­sider \(hnh^{-1} N + hN\). Since the quo­tient is well-defined, this is \((hnh^{-1}h) N\), which is \(hnN\) or \(hN\) (since \(nN = N\), be­cause \(N\) is a sub­group of \(G\) and hence is closed un­der the group op­er­a­tion). But that means \(hnh^{-1}N\) is the iden­tity el­e­ment of the quo­tient group, since when we added it to \(hN\) we ob­tained \(hN\) it­self.

That is, \(hnh^{-1}N = N\). There­fore \(hnh^{-1} \in N\).

Since this rea­son­ing works for any \(h \in G\), it fol­lows that \(N\) is closed un­der con­ju­ga­tion by el­e­ments of \(G\), and hence is nor­mal.


  • Normal subgroup

    Nor­mal sub­groups are sub­groups which are in some sense “the same from all points of view”.